Q1. What is the probability of rolling a sum of 12 on three rolls of six-sided dice? Express your answer as a decimal number only. Show your R code.

#define the possible outcomes of a singe die throw, find every combination of 3 dice throws, and find the total sample size
six_die = c(1,2,3,4,5,6) 
combo = expand.grid(roll1 = six_die, roll2 = six_die, roll3 = six_die) 
samplesSize <- nrow(combo)

#sum up the results of the 3 throws for each case and find how many resulted in a sum of 12
sum <- rowSums(combo) 
eventSize <- length(sum[sum==12])

#divide the amount of times the sum was 12 by the total number of samples taken
eventSize/samplesSize 
## [1] 0.1157407

The odds of getting a total sum of 12 when rolling 3 dice is 11.57%. Of the 216 possible outcomes, 25 of them have a sum of 12.

Q2. A newspaper company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of customers. The data is summarized in the table below.

What is the probability that a customer is male and lives in ‘Other’ or is female and lives in ‘Other’? Express your answer as a decimal number only. Show your R code.

#create a martix to match the information from the above table.
customerData = matrix(
  c(200, 300, 200, 100, 100, 200, 200, 100, 200, 100), 
  nrow = 5,   
  ncol = 2,         
  byrow = TRUE          
)
colnames(customerData) <- c("Male", "Female")
rownames(customerData) <- c("Apartment", "Dorm", "With Parent(s)", "Sorority/Fraternity House", "Other")
customerData
##                           Male Female
## Apartment                  200    300
## Dorm                       200    100
## With Parent(s)             100    200
## Sorority/Fraternity House  200    100
## Other                      200    100
#find row and column totals, bind results to matrix, and update names for clarity
customerData <- cbind(customerData, rowSums(customerData))
customerData <- rbind(customerData, colSums(customerData))
colnames(customerData) <- c("Male", "Female", "Total")
rownames(customerData) <- c("Apartment", "Dorm", "With Parent(s)", "Sorority/Fraternity House", "Other", "Total")
customerData
##                           Male Female Total
## Apartment                  200    300   500
## Dorm                       200    100   300
## With Parent(s)             100    200   300
## Sorority/Fraternity House  200    100   300
## Other                      200    100   300
## Total                      900    800  1700
#divide all entries by the total sample size to get the probability of each outcome
proportionalCustomerData <- customerData/customerData[6,3]
proportionalCustomerData
##                                 Male     Female     Total
## Apartment                 0.11764706 0.17647059 0.2941176
## Dorm                      0.11764706 0.05882353 0.1764706
## With Parent(s)            0.05882353 0.11764706 0.1764706
## Sorority/Fraternity House 0.11764706 0.05882353 0.1764706
## Other                     0.11764706 0.05882353 0.1764706
## Total                     0.52941176 0.47058824 1.0000000

The probability that one is male and lives in other is 11.76%. The probability that one is female and lives in other is 5.88%.

Q3. Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a diamond for the second card drawn, if the first card, drawn without replacement, was a diamond? Express your answer as a decimal number only. Show your R code.

We are looking for \(P(D2|D1)\), or the probability that the second card drawn is a diamond given that the first card was also a diamond. We can solve this in R with the code below.

pd1 = 13/52 # 13 diamond cards out of a total of 52
pd1
## [1] 0.25
pd2_given_d1 = 12/51 #after the initial draw, there is now 1 less diamond card in particular, and 1 less card in general
pd2_given_d1
## [1] 0.2352941

This result makes intuitive sense. If you draw a diamond card first, your odds of doing so again should be less than before.

Q4.A coordinator will select 10 songs from a list of 20 songs to compose an event’s musical entertainment lineup. How many different lineups are possible? Show your R code.

Assuming that the coordinator cares about order, and that they would not want to repeat songs, we can use the formula \(\displaystyle \frac{n!}{(n-r)!}\) to describe the number of possible permutations. The amount of songs available is n=20; and we will take r=10 songs.

n <- 20
r <- 10
factorial(n)/factorial(n-r)
## [1] 670442572800

Q5. You are ordering a new home theater system that consists of a TV, surround sound system, and DVD player. You can choose from 20 different TVs, 20 types of surround sound systems, and 18 types of DVD players. How many different home theater systems can you build? Show your R code.

Now we are looking to solve for how many combinations we can make from these 3 items. Here the order does not matter, but what is sampled does. Since we are only taking 1 of each item, we know there are only 20 combinations of TVs, 20 of sound systems, and 18 of DVD players. We can find the overall number of combinations simply by multiplying these 3 values together.

tv <- 20
sound <- 20
dvd <- 18
tv*sound*dvd
## [1] 7200

Q6. A doctor visits her patients during morning rounds. In how many ways can the doctor visit 10 patients during the morning rounds? Show your R code.

Again we are trying to solve for possible permutations without reputation, so we’ll use the formula \(\displaystyle \frac{n!}{(n-r)!}\) like before.

n <- 10
r <- 10
factorial(n)/factorial(n-r)
## [1] 3628800

Q7. If a coin is tossed 7 times, and then a standard six-sided die is rolled 3 times, and finally a group of four cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible? Show your R code

Here we will solve for 3 different combinations, 2 with replacement, one without, and then multiply those results by each other to get the total number of combinations. I am assuming that order does not matter for this problem, if it did then the numbers would be much higher.

\(\displaystyle \frac{(r+n-1)!}{r!(n-1)!}\) - Combinations w/repeats (used for the 7 coin flips and 3 dice rolls)

\(\displaystyle \frac{n!}{r!(n-r)!}\) - Combinations w/o repeats (used for the 4 card draws)

coin_combos <- factorial(7+2-1)/(factorial(7)*factorial(2-1))
dice_combos <- factorial(3+6-1)/(factorial(3)*factorial(6-1))
card_combos <- factorial(52)/(factorial(4)*factorial(52-4))
coin_combos*dice_combos*card_combos
## [1] 121284800

Q8. In how many ways may a party of four women and four men be seated at a round table if the women and men are to occupy alternate seats. Show your R code.

This problem comes down to a problem of permutation without replacement. However we can’t just use the basic formula due to the requirement that men and women must sit in alternate seats. We can however solve it by treating it as 8 sequential permutations each with a single sample from the alternating groups. The first 2 seat selections both can be written as:

\(\displaystyle \frac{n!}{(n-r)!} = \displaystyle \frac{4!}{(4-1)!} = 4\)

The next 2 seat selections, now pulling from a set of 3 men and 3 women can both be written as:

\(\displaystyle \frac{n!}{(n-r)!} = \displaystyle \frac{3!}{(3-1)!} = 3\)

And so on. The final result is solved in the following line of R code:

(4*4)*(3*3)*(2*2)*(1*1)
## [1] 576

Q9. An opioid urinalysis test is 95% sensitive for a 30-day period, meaning that if a person has actually used opioids within 30 days, they will test positive 95% of the time P( + |User) =.95. The same test is 99% specific, meaning that if they did not use opioids within 30 days, they will test negative P( - | Not User) = .99. Assume that 3% of the population are users. Then what is the probability that a person who tests positive is actually a user P(User | +)? Show your R code.

Lets begin by writing out some of the given probabilities:

\(P(User) = 0.03, P(+|User) = 0.95, P(-|Not User) = 0.99, P(User|+) = ?\)

Note: We can use the information \(P(-|Not User) = 0.99\) to infer that \(P(+|Not User) = 0.01\)

We can solve this using Bayes Theorem:

\(P(User|+) = \displaystyle \frac{P(+|User)P(User)}{P(+)}\)

The numerator is easy to solve using the given values:

num <- 0.95*0.03
num
## [1] 0.0285

While the denominator \(P(+)\) is a bit more complicated. We must add up the the probability of every outcome where the test is positive.

denom <- 0.95*0.03 + 0.01*0.97
denom
## [1] 0.0382

With Bayes Theorem complete, we can solve for \(P(User|+)\)

num/denom
## [1] 0.7460733

That is to say that if you test positive for opioid use, there is still only a 74.6% chance that you are actually a user.

Q10. You have a hat in which there are three pancakes. One is golden on both sides, one is brown on both sides, and one is golden on one side and brown on the other. You withdraw one pancake and see that one side is brown. What is the probability that the other side is brown? Explain.

Note: we can write conditional probability in a more general form \(P(A|B) = \displaystyle \frac{P(A*B)}{P(B)}\)

We are looking for the conditional probability \(P(B2|B1)\). I began this problem by solving for \(P(B2*B1)\) or the chance that you pulled the pancake with 2 brown edges. We can easily see that this is 1/3rd.

p_b2_and_b1 <- 1/3

Now if we can find \(P(B1)\) we should be able to solve for the conditional probability. Thankfully this is also rather simple, as there are 3 pancake sides that are brown, and 3 that are golden, meaning that there a 50% chance of getting a brown side first.

p_b1 = 1/2

Plugging into the conditional probability formula gets us the following result:

\(\displaystyle \frac{P(B2*B1)}{P(B1)}\)

p_b2_and_b1/p_b1
## [1] 0.6666667

This makes some intuitive sense, as of the 3 brown sides in the hat, 2 of them are opposite another brown side. So if you pull one of them, you have a 2/3rds chance that the other side will be brown as well.

\(P(B2|B1) = 2/3\)